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Find the volume of the solid formed by rotating the region enclosed by the curve of \( f \) and the x-axis about the x-axis through 360 degrees.

Answer :

The volume of the solid generated by rotating the region enclosed by the curve of $f(x) = x^2 - 1$ and the x-axis around the x-axis is $\frac{16\pi}{15}$.

To find the volume of the solid generated by rotating the region enclosed by the curve of a function $f(x)$ and the x-axis around the x-axis, we use the formula:

$V = \int_{a}^{b} \pi [f(x)]^2 dx$

where $a$ and $b$ are the limits of integration.

Let's assume the function $f(x) = x^2 - 1$, then we have:

$V = \int_{-1}^{1} \pi [(x^2 - 1)]^2 dx$

$V = \int_{-1}^{1} \pi (x^4 - 2x^2 + 1) dx$

$V = \pi \left[\frac{x^5}{5} - \frac{2x^3}{3} + x\right]_{-1}^{1}$

$V = \pi \left[\left(\frac{1}{5} - \frac{2}{3} + 1\right) - \left(-\frac{1}{5} + \frac{2}{3} - 1\right)\right]$

$V = \pi \left[\frac{16}{15}\right]$

$V = \frac{16\pi}{15}$

Therefore, the volume of the solid generated by rotating the region enclosed by the curve of $f(x) = x^2 - 1$ and the x-axis around the x-axis is $\frac{16\pi}{15}$.

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